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There is a proof on wikipedia : https://groupprops.subwiki.org/wiki/A5_is_simple

I was searching of proofs that $A_5$ is simple using conjugacy classes argument as I have recently read the chapter on sylow theorems and simple groups which have theory on conjugacy classes.

But I am unable to follow the proof given in the link:

Why the conjugacy classes size is only $1,12, 20, 15$.

I think as $\mbox{cl}(a)$ divides $|G|$, where $\mbox{cl}(a)$ is conjugacy class of $a$ on $G$. It can be $1,2,3,4,5,6,10,12,15,20,30,60$.

Am I wrong ?

"A normal subgroup must contain the conjugacy class of size $1$, and one or more other conjugacy classes"

Is the statement written just above due to following reason?: If $N$ is a normal subgroup and let $n\in \mathbb N$ be any of it's elementa then $gng^{-1}$ ( $g \in G$) lies in $N$ but it is not necessarily $n$ itself (One of the elements it is true when $g = e$(identity)).

"Thus, the order of any normal subgroup must be a sum of some of these numbers, including the $1$. By Lagrange's theorem, the order must also divide the order of the group."

I understand the statement written just above.

"But no such sum among these numbers divides $60$, other than $1$ and $60$ themselves."

I don't understand it this line.

Can you please help with the questions in above proof ?

Sarvesh Ravichandran Iyer
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  • Can you at least prove that a conjugacy class cannot have size $60$? – Derek Holt Sep 12 '21 at 12:57
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    Hint i: the conjugacy classes have sizes 1,12,12,15,20. The page you reference gives links to a proof of this, though it's hard to chase through the links. Hint ii: [the last bit] $N$ has to have the identity, suppose it also contains the class of size 20; well 21 doesn't divide 60 so there's more; but that takes us beyond 30 so we must hit 60. So $N$ doesn't have the class of size 20; suppose it has the class of size 15, well 16 doesn't divide so there's more etc etc etc. – ancient mathematician Sep 12 '21 at 15:02
  • Comment: "due to this reason" : what you write is garbled. Slow down. Separate into small sentences. – ancient mathematician Sep 12 '21 at 15:05
  • Hint: The sizes of the conjugacy classes come from combining two pieces of information. A) The sizes of the conjugacy classes of $S_5$. These can be calculated by combinatorial arguments as we know that two permutations in $S_5$ are conjugate if and only if they have the same cycle type. B) How the $S_5$-conjugacy classes of even permutations "split" in $A_5$. That gives you the sizes. Then follow the program outlined by ancient mathematician. – Jyrki Lahtonen Sep 16 '21 at 18:03
  • @ancientmathematician I have edited it 2 days ago and made it clear. Hope it is fine now. –  Sep 19 '21 at 06:43
  • @DerekHolt Well I was unable to. I assumed that let conjugacy class of an element a is 60 elements. so for every element $x\in A_5$ there exists some $g\in A_5$ such that $x= g a g^{-1}$.But I am unable to get any contradiction. –  Sep 19 '21 at 06:51
  • I don't think it's clearly written at all. "If $n\in N$ is a normal subgroup"?! "$g\in G$ is the order of the group"?! And why do you think that $g^{-1}ng=n$ only happens for $g=e$? Until you sort all this out for yourself it's pointless to try to tackle a serious problem. – ancient mathematician Sep 19 '21 at 06:51
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    @ancientmathematician I am really sorry for inconvinience. I am editing it more –  Sep 19 '21 at 06:52
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    @ancientmathematician I have edited it again. Kindly have a look if you have some spare time. –  Sep 19 '21 at 06:57
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    @ancientmathematician Your first comment: "suppose it also contains the class of size 20; well 21 doesn't divide 60 " Why must 21 divide 60. I think 20 dividing 60 must be the only condition in it. –  Sep 19 '21 at 06:59
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    If every element is conjugate to $g$, then $e$ is conjugate to $g$. But $xex^{-1}=e$ for every $x\in G$. No conjugacy class can have the size of the group, unless the group has order $1$. You say you were unable to get a contradiction from the assumption that the conjugacy class had 60 elements. That seems... hard to credit. – Arturo Magidin Sep 19 '21 at 07:03
  • WE are building a normal subgroup: it has to contain $e$ that's one element; suppose it contains the ccl of $20$ elements; could we stop there? only if $21|60$, which it doesn't; so there are more elements; and in no time $|N|=60$. – ancient mathematician Sep 19 '21 at 07:05
  • The edit helps. But you have left the mathematically wrong statement that $g^{-1}ng$ is not $n$. – ancient mathematician Sep 19 '21 at 07:08
  • Oh dear, the subwiki article is a stub. I think if someone writes it up really neatly here, we can push whatever's written here into the subwiki article. I will try to be as clear as possible, but first let me see if this question is clear to me or not! – Sarvesh Ravichandran Iyer Sep 19 '21 at 09:59
  • @TeresaLisbon I have removed the issues by editing and please help if you have some spare time –  Sep 19 '21 at 11:48
  • @ArturoMagidin Thanks! –  Sep 19 '21 at 11:50
  • @Avenger I think cl(a) already contains e.The definition of conjugacy class is "conjugacy class of a is the set cl(a) = {$xax^{-1} | x \in G$}." So, why do you think that x $\in $G in the cl(a) doesn't contains e. –  Sep 19 '21 at 12:01
  • @Avenger Thanks, I find this to be good enough. I would encourage potential answerers to write their answer , not only addressing this question, but recognizing that the wiki article is a stub, so hopefully, I'll try to find out how I can edit that article by using a good answer below, with due credit afforded. – Sarvesh Ravichandran Iyer Sep 19 '21 at 14:54
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    Note : I edited this question to add in some MathJax and correct a spelling error, but even before that , I found it an acceptable post for the reasons I've outlined above. +1. – Sarvesh Ravichandran Iyer Sep 19 '21 at 15:09
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    @TeresaLisbon Thanks a lt. That's very kind of you. –  Sep 19 '21 at 15:34
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    @Avenger Thanks a lot, for placing bounty on this question and for providing the context. As for the question not getting attention, it's happened with me as well, I've had some questions that even went through a bounty without getting any attention (after seven days of being bountied, it had only 45 views!) so I suppose it's something really arbitrary. Placing the bounty always helps, though! – Sarvesh Ravichandran Iyer Sep 19 '21 at 15:37
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    @Avenger: "Why do you think the conjugacy class does not contain $e$?" Your use of language is extremely poor. The conjugacy class of $a$ doesn't contain the $x$s that you use to conjugate, it contains the results of conjugation. If $e=xax^{-1}$, then $a=e$. The only conjugacy class that "contains" $e$ is ${e}$, the conjugacy class of $e$, which ontains $e$ and nothing else. The fact that we use $e$ as among the conjugating elements guarantees that the conjugacy class of $a$ contains $a$, not that it "contains $e$". If this is the kind of confusion you have. you need to review. – Arturo Magidin Sep 19 '21 at 16:24
  • @Arturo Magidin English is not my native language –  Sep 20 '21 at 04:30
  • @ArturoMagidin I am very sorry for the inconvenience caused. I am here to learn and causing incovenience is never my intention. I will be more careful while framing questions from now on. –  Sep 20 '21 at 04:38
  • @Avenger: English is my fourth language. So personally, I don't find that as a particularly good explanation. – Arturo Magidin Sep 20 '21 at 12:48
  • @ArturoMagidin I understood your this comment "completely:hy do you think the conjugacy class does not contain e?" Your use of language is extremely poor. The conjugacy class of a doesn't contain the xs that you use to conjugate, it contains the results of conjugation. If e=xax−1, then a=e." But how exactly should I prove that |N| will be no less than 60. –  Sep 22 '21 at 05:03
  • @ancientmathematician Your comment :"WE are building a normal subgroup: it has to contain e that's one element; suppose it contains the ccl of 20 elements; could we stop there? only if 21|60, which it doesn't; so there are more elements; and in no time |N|=60". I got it but can you rigoriously prove that |N| cannot be less than 60? Your idea of using 21 doesn't dividing 60 and similarly is very informative. –  Sep 22 '21 at 05:09

2 Answers2

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I attempt an elementary argument.

Before starting notice this. In any group $G$ the number of conjugates of an element $x$ is $|G:C(x)|$.

[Proof: We need to count coincidences among the $g^{-1}xg$ as $g\in G$. Well $g_1^{-1}xg_1 = g_2^{-1}xg_2$ if and only if $(g_1 g_2^{-1})^{-1}x(g_1 g_2^{-1})=x$; that is, if and only if $(g_1 g_2^{-1})\in C(x)$. Hence all the elements of $C(x)g$ conjugate $x$ to the same element, and that's the only way coincidences happen.]

Any permutation in $S_5$ can be expressed as a product of disjoint cycles, and so has one of the following forms: $$ (abcde), (abcd)(e), (abc)(de), (abc)(d)(e), (ab)(cd)(e), (ab)(c)(d)(e) (a)(b)(c)(d)(e). $$

Of these the only ones which are even and so lie in $A_5$ are $$ (abcde), (abc)(d)(e), (ab)(cd)(e), (a)(b)(c)(d)(e) $$ whose orders (as listed) are $$ 5,3,2,1. $$ The number of elements of each type is $$ \frac{5!}{5}=24, \frac{5!}{3\cdot 2!}=20, \frac{5!}{2\cdot 2 \cdot 1 \cdot 2!}=15, \frac{5!}{5!}=1. $$

Now let us split these into conjugacy classes.

(i) As there are no elements of order $10$ or $15$ the order of $C((abcde))$ is (by Lagrange) $5$, and so each of the $24$ $5$-cycles is conjugate to $|A_5:C((abcde))|=12$ others. That is, there are two conjugacy classes of size $12$.

(ii) As there are no elements of order $6$ or $15$ the order of $C((abc)(d)(e))$ is (by Lagrange) $3$, and so each of the $20$ $3$-cycles is conjugate to $|A_5:C((abc)(d)(e))|=20$ others. That is, there is one conjugacy classes of size $20$.

(iii) As there are no elements of order $6$ or $10$ the order of $C((ab)(cd)(e))$ is (by Lagrange) $2$ or $4$, so that each of the $15$ double transpositions $(ab)(cd)(e)$ is conjugate to $|A_5:C((ab)(cd)(e))|$ elements, that is $30$ or $15$ elements. Clearly it is the latter case which holds and there is one conjugacy class of size $15$.

(iv) The identity $(a)(b)(c)(d)(e)$ is in a class of size $1$.

Now let $N$ be a normal subgroup of $A_5$. Since $n\in N$ implies every conjugate $g^{-1}ng\in N$ we have that $N$ is a disjoint union of pieces whose sizes are $1,15,20,12,12$.

Suppose $N\ne A_5$, $N\ne \{(a)(b)(c)(d)(e)\}$. By Lagrange its order divides $60$ and is strictly less than $60$. Also, $N$ must contain the identity and at least one other conjugacy class so its size is at least $13$. The only candidate orders are then $15,20,30$. As $N$ must contain the class of the identity none of these can be built from pieces of the given sizes.

Hence the only normal subgroups of $A_5$ are the identity subgroup and the whole group.

ancient mathematician
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  • "order of C((abcde))" is 5 using Lagrange theorem. How exactly? C(abcde) is a subgroup of $A_5$ which means that it's order divides 60. but there is no element of order 10,15 or 30 but now how did you deduced that it's order must be 5? Can you please tell? –  Oct 15 '21 at 06:42
  • @Avenger If $y$ of order $p$ centralises $x$ of order $q$ where $p,q$ are coprime then the order of $xy$ is $pq$. Now clearly $C(x)$ (with $x=(abcde)$) has order divisible by $5$. If it were $10$ then since there are no elements of order $10$ or $15$ we'd have a group of order $10$ all of whose elements commute with $x$ and have order $5$. Take $t$ as one of these not a power of $x$. Then $x^i t^j$ must all be different and $C(x)$ has order $25$. [Or more easily use Cauchy's Thm on the existence of elements of order $p$ - I am avoiding that.] The other cases similarly. – ancient mathematician Oct 15 '21 at 07:05
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You need to know about cycle structures. $A_5$ is the group of all even permutations on $5$ elements. These look like this:

  1. $()$ - the identity
  2. $(a\ b)(c\ d)$
  3. $(a\ b\ c)$
  4. $(a\ b\ c\ d\ e)$

All elements are of the forms above. Moreoever, conjugatation preserves cycle structure. That is, $g$ and $h^{-1}gh$ have the same cycle structure (e.g. if $g$ is a $3$-cycle, then $h^{-1}gh$ is also a $3$-cycle.)

Now, if you accept the statements above, you can accept that the conjugacy classes of $A_5$ are just the sets of different cycle types and you can count them:

  1. cycle type $()$ - just one
  2. $3$-cycles - $20$ in total
  3. $(a\ b)(c\ d)$-cycles - $15$ in total
  4. This case is a bit more complicated.

This is why the proof you use can immediately discount all the other potential sizes for the conjugacy classes and only consider $1,12,15,20$.

The more complicated case is because $5$-cycles split into two conjugacy classes in $A_5$, so rather being size $24$, there are two classes of size $12$. You should find in your textbook or the internet results on cycle structures in $S_n$ and $A_n$, they will give you more details.

This might help you: pp37-39, http://dec41.user.srcf.net/notes/IA_M/groups.pdf

Teddy38
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